Question

Substance A has atomic mass number $16$ and half life of $1$ day. Another substance $B$ has atomic mass number $32$ and half life of $\frac{1}{2}$ day. If both $A$ and $B$ simultaneously start undergo radio activity at the same time with initial mass $320g$ each, how many total atoms of $A$ and $B$ combined would be left after 2 days.

• A. $3.38\times 10^{24}$
• B. $6.76\times 10^{24}$
• C. $6.76\times 10^{23}$
• D. $1.69\times 10^{24}$

Answer 1

Answer: (A)

$\left(N_0\right)A=\frac{320}{16}=20\text{ moles }$
$\left(N_0\right)B=\frac{320}{32}=10\text{ moles }$
$N_A=\frac{{\left(N_0\right)}_A}{(2{)}^{2/1}}=\frac{20}{4}=5$
$N_B=\frac{{\left(N_0\right)}_B}{(2{)}^{2/5}}=\frac{10}{2^4}=0.625$
$\text{ Total }N=5.625$
$\text{ No. of atoms }=5.625\times 6.023\times 10^{23}$
$=3.38\times 10^{24}$